Question

By using properties of determinants,show that:I$\begin{vmatrix} x & x^2 & yz\\\ y & y^2 & zx\\\z&z^2&xy \end{vmatrix}$=(x-y)(y-z)(z-x)(xy+yz+zx)

Answer 1

Let △=$\begin{vmatrix} x & x^2 & yz\\\ y & y^2 & zx\\\z&z^2&xy \end{vmatrix}$

Applying R2 → R2 − R1 and R3 → R3 − R1, we have:

△=$\begin{vmatrix} x & x^2 & yz\\\ y-x & y^2-x^3 & zx-yz\\\z-x&z^2-x^2&xy-yz \end{vmatrix}$

=$\begin{vmatrix} x & x^2 & yz\\\\- (x-y) & -(x-y)(x+y)& zx-yz\\\z-x&z^2-x^2&-y(z-x) \end{vmatrix}$

=(x-y)(z-x)$\begin{vmatrix} x & x^2 & yz\\\ -1 &-x-y & z\\\1&z+x&-y \end{vmatrix}$

Applying R3 → R3 + R2, we have:

△=(x-y)(z-x)$\begin{vmatrix} x & x^2 & yz\\\ -1 &-x-y & z\\\0&z-y&z-y \end{vmatrix}$

=(x-y)(z-x)(z-y)$\begin{vmatrix} x & x^2 & yz\\\ -1 &-x-y & z\\\0&1&1 \end{vmatrix}$

Expanding along R3, we have:

△=[(x-y)(z-x)(z-y)]\bigg[(-1)$$\begin{vmatrix} x &yz \\\ -1 & z \end{vmatrix}+1$\begin{vmatrix} x &x^2 \\\ -1 & -x-y \end{vmatrix}$ $\bigg]$
=[(x-y)(z-x)(z-y)][(-xz-yz)+(-x2-xy+x2)]
=(x-y)(y-z)(z-x)(xy+yz+zx)
Hence, the given result is proved.